Therefore the output switches back to 0V and the circuit is back in the starting state. ![]() C1 very quickly discharges through the discharge transistor making the voltage across C1 0V. This results in the flip flop output being on and therefore turn on the discharge transistor. When this happens COMP2’s output will turn on which will in turn reset the flip flop. Since the inverting input of COMP2 is connected to 2/3 VCC and the non inverting input to the capacitor the non inverting input voltage will become greater than the inverting input. Note that the capacitor is connected to the threshold input on the 555 which is internally connected to the non-inverting input to COMP2.Įventually the voltage across C1 will become larger than 2/3 VCC. With the discharge pin no longer discharging the capacitor C1 begins to charge through R3 and the potentiometer. As the flip-flop output is 0V the output stage of the 555 inverts this and therefore the output switches to 9V. Even when the switch is depressed the transistor stays off because the flip flop remembers what the last input was (i.e. ![]() The flip flop output is connected to the base of the discharge transistor so when the switch is pressed the transistor stops discharging the capacitor. This short pules from COMP1 sets the flip flop and therefore the output of the flip flop will be 0V. Therefore the output of COMP1 is on for a short amount of time. ![]() This means that normally the output of COMP1 is off but when the switch is pressed, the inverting pin (for a short time), is at a lower voltage than the non-inverting pin. COMP1 inverting input is connected to the trigger pin and the non-inverting input is connected to 1/3VCC. This is done by pressing the switch (when the switch contacts are shorted the trigger input is connected directly to ground and therefore will have a voltage of 0V). To start the monostable the trigger voltage needs to quickly be dropped to 0V.
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